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How To Match the begining and the end of string

How To Match the begining and the end of a string but ignore everything in between. I made it like this, but the result is always zero for everything.
document.write( (str.match(/^["“”'’]$/g) || []).length );
// expected results
str = "please \"don't\" talk"; // 2
str = "please don't talk"; // 0
str = "Thomas' car"; // 1
to macth all the qoutes like in my code str.match(/["“”'’]/g)
but giving ^ and $ is always giving zero result

Try
document.write( (str.match(/["“”'’]\s|\s["“”'’]/g) || []).length );
and you will get a count for every quote that has an adjacent space. It won't work for a quote at the end of the string. If you need that also, ask.
The | separates different searches in this case. So we are searching for a quote followed by a space or a space followed by a quote. If you explicitly want the \" to be counted then the expression would be a little more complex. I am assuming the \ is just escaping " in the string.

You need
\B["“”'’]|["“”'’]\B
See demo
The non-word boundary \B make sure the match only occurs at a position that is not word boundary position:
\B matches at any position between two word characters as well as at any position between two non-word characters.
So, \B["“”'’] matches all substrings that are not preceded with a word ([a-zA-Z0-9_]) character, and ["“”'’]\B matches the apostrophes that are not followed by a word character.

Regex to match # followed by square brackets containing a number

I want to parse a pattern similar to this using javascript:
#[10] or #[15]
With all my efforts, I came up with this:
#\\[(.*?)\\]
This pattern works fine but the problem is it matches anything b/w those square brackets. I want it to match only numbers. I tried these too:
#\\[(0-9)+\\]
and
#\\[([(0-9)+])\\]
But these match nothing.
Also, I want to match only pattern which are complete words and not part of a word in the string. i.e. should contain spaces both side if its not starting or ending the script. That means it should not match phrase like this:
abxdcs#[13]fsfs
Thanks in advance.

Use the regex:
/(?:^|\s)#\[([0-9]+)\](?=$|\s)/g
It will match if the pattern (#[number]) is not a part of a word. Should contain spaces both sides if its not starting or ending the string.
It uses groups, so if need the digits, use the group 1.
Testing code (click here for demo):
console.log(/(?:^|\s)#\[([0-9]+)\](?=$|\s)/g.test("#[10]")); // true
console.log(/(?:^|\s)#\[([0-9]+)\](?=$|\s)/g.test("#[15]")); // true
console.log(/(?:^|\s)#\[([0-9]+)\](?=$|\s)/g.test("abxdcs#[13]fsfs")); // false
console.log(/(?:^|\s)#\[([0-9]+)\](?=$|\s)/g.test("abxdcs #[13] fsfs")); // true
var r1 = /(?:^|\s)#\[([0-9]+)\](?=$|\s)/g
var match = r1.exec("#[10]");
console.log(match[1]); // 10
var r2 = /(?:^|\s)#\[([0-9]+)\](?=$|\s)/g
var match2 = r2.exec("abxdcs #[13] fsfs");
console.log(match2[1]); // 13
var r3 = /(?:^|\s)#\[([0-9]+)\](?=$|\s)/g
var match3;
while (match3 = r3.exec("#[111] #[222]")) {
console.log(match3[1]);
}
// while's output:
// 111
// 222

You were close, but you need to use square brackets:
#\[[0-9]+\]
Or, a shorter version:
#\[\d+\]
The reason you need those slashes is to "escape" the square bracket. Usually they are used for denoting a "character class".
[0-9] creates a character class which matches exactly one digit in the range of 0 to 9. Adding the + changes the meaning to "one or more". \d is just shorthand for [0-9].
Of course, the backslash character is also used to escape characters inside of a javascript string, which is why you must escape them. So:
javascript
"#\\[\\d+\\]"
turns into:
regex
#\[\d+\]
which is used to match:
# a literal "#" symbol
\[ a literal "[" symbol
\d+ one or more digits (nearly identical to [0-9]+)
\] a literal "]" symbol
I say that \d is nearly identical to [0-9] because, in some regex flavors (including .NET), \d will actually match numeric digits from other cultures in addition to 0-9.

You don't need so many characters inside the character class. More importantly, you put the + in the wrong place. Try this: #\\[([0-9]+)\\].

regex string replace

I am trying to do a basic string replace using a regex expression, but the answers I have found do not seem to help - they are directly answering each persons unique requirement with little or no explanation.
I am using str = str.replace(/[^a-z0-9+]/g, ''); at the moment. But what I would like to do is allow all alphanumeric characters (a-z and 0-9) and also the '-' character.
Could you please answer this and explain how you concatenate expressions.

This should work :
str = str.replace(/[^a-z0-9-]/g, '');
Everything between the indicates what your are looking for
/ is here to delimit your pattern so you have one to start and one to end
[] indicates the pattern your are looking for on one specific character
^ indicates that you want every character NOT corresponding to what follows
a-z matches any character between 'a' and 'z' included
0-9 matches any digit between '0' and '9' included (meaning any digit)
- the '-' character
g at the end is a special parameter saying that you do not want you regex to stop on the first character matching your pattern but to continue on the whole string
Then your expression is delimited by / before and after.
So here you say "every character not being a letter, a digit or a '-' will be removed from the string".

Just change + to -:
str = str.replace(/[^a-z0-9-]/g, "");
You can read it as:
[^ ]: match NOT from the set
[^a-z0-9-]: match if not a-z, 0-9 or -
/ /g: do global match
More information:
https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Regular_Expressions

Your character class (the part in the square brackets) is saying that you want to match anything except 0-9 and a-z and +. You aren't explicit about how many a-z or 0-9 you want to match, but I assume the + means you want to replace strings of at least one alphanumeric character. It should read instead:
str = str.replace(/[^-a-z0-9]+/g, "");
Also, if you need to match upper-case letters along with lower case, you should use:
str = str.replace(/[^-a-zA-Z0-9]+/g, "");

str = str.replace(/\W/g, "");
This will be a shorter form

We can use /[a-zA-Z]/g to select small letter and caps letter sting in the word or sentence and replace.
var str = 'MM-DD-yyyy'
var modifiedStr = str.replace(/[a-zA-Z]/g, '_')
console.log(modifiedStr)

How do I combine these two regular expressions into one?

I'm writing a rudimentary lexer using regular expressions in JavaScript and I have two regular expressions (one for single quoted strings and one for double quoted strings) which I wish to combine into one. These are my two regular expressions (I added the ^ and $ characters for testing purposes):
var singleQuotedString = /^'(?:[^'\\]|\\'|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*'$/gi;
var doubleQuotedString = /^"(?:[^"\\]|\\"|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*"$/gi;
Now I tried to combine them into a single regular expression as follows:
var string = /^(["'])(?:[^\1\\]|\\\1|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*\1$/gi;
However when I test the input "Hello"World!" it returns true instead of false:
alert(string.test('"Hello"World!"')); //should return false as a double quoted string must escape double quote characters
I figured that the problem is in [^\1\\] which should match any character besides matching group \1 (which is either a single or a double quote - the delimiter of the string) and \\ (which is the backslash character).
The regular expression correctly filters out backslashes and matches the delimiters, but it doesn't filter out the delimiter within the string. Any help will be greatly appreciated. Note that I referred to Crockford's railroad diagrams to write the regular expressions.

You can't refer to a matched group inside a character class: (['"])[^\1\\]. Try something like this instead:
(['"])((?!\1|\\).|\\[bnfrt]|\\u[a-fA-F\d]{4}|\\\1)*\1
(you'll need to add some more escapes, but you get my drift...)
A quick explanation:
(['"]) # match a single or double quote and store it in group 1
( # start group 2
(?!\1|\\). # if group 1 or a backslash isn't ahead, match any non-line break char
| # OR
\\[bnfrt] # match an escape sequence
| # OR
\\u[a-fA-F\d]{4} # match a Unicode escape
| # OR
\\\1 # match an escaped quote
)* # close group 2 and repeat it zero or more times
\1 # match whatever group 1 matched

This should work too (raw regex).
If speed is a factor, this is the 'unrolled' method, said to be the fastest for this kind of thing.
(['"])(?:(?!\\|\1).)*(?:\\(?:[\/bfnrt]|u[0-9A-F]{4}|\1)(?:(?!\\|\1).)*)*/1
Expanded
(['"]) # Capture a quote
(?:
(?!\\|\1). # As many non-escape and non-quote chars as possible
)*
(?:
\\ # escape plus,
(?:
[\/bfnrt] # /,b,f,n,r,t or u[a-9A-f]{4} or captured quote
| u[0-9A-F]{4}
| \1
)
(?:
(?!\\|\1). # As many non-escape and non-quote chars as possible
)*
)*
/1 # Captured quote

Well, you can always just create a larger regex by just using the alternation operator on the smaller regexes
/(?:single-quoted-regex)|(?:double-quoted-regex)/
Or explicitly:
var string = /(?:^'(?:[^'\\]|\\'|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*'$)|(?:^"(?:[^"\\]|\\"|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*"$)/gi;
Finally, if you want to avoid the code duplication, you can build up this regex dynamically, using the new Regex constructor.
var quoted_string = function(delimiter){
return ('^' + delimiter + '(?:[^' + delimiter + '\\]|\\' + delimiter + '|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*' + delimiter + '$').replace(/\\/g, '\\\\');
//in the general case you could consider using a regex excaping function to avoid backslash hell.
};
var string = new RegExp( '(?:' + quoted_string("'") + ')|(?:' + quoted_string('"') + ')' , 'gi' );

Regular expression negative match

I can't seem to figure out how to compose a regular expression (used in Javascript) that does the following:
Match all strings where the characters after the 4th character do not contain "GP".
Some example strings:
EDAR - match!
EDARGP - no match
EDARDTGPRI - no match
ECMRNL - match
I'd love some help here...

Use zero-width assertions:
if (subject.match(/^.{4}(?!.*GP)/)) {
// Successful match
}
Explanation:
"
^ # Assert position at the beginning of the string
. # Match any single character that is not a line break character
{4} # Exactly 4 times
(?! # Assert that it is impossible to match the regex below starting at this position (negative lookahead)
. # Match any single character that is not a line break character
* # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
GP # Match the characters “GP” literally
)
"

You can use what's called a negative lookahead assertion here. It looks into the string ahead of the location and matches only if the pattern contained is /not/ found. Here is an example regular expression:
/^.{4}(?!.*GP)/
This matches only if, after the first four characters, the string GP is not found.

could do something like this:
var str = "EDARDTGPRI";
var test = !(/GP/.test(str.substr(4)));
test will return true for matches and false for non.